SPM Mathematics 2019, Paper 2 (Question 7 & 8)

Question 7:
Diagram 3 shows the straight lines PR, SQ, QT and ST drawn on a Cartesian plane. It is given that PR and ST are parallel and SQ = QT – 10 units.

Diagram 3

Find the equation of the straight line PR. [4 marks]

Solution:

$\begin{array}{l} Q = (x, 8) \\ S Q = 10 \\ \sqrt{{[x - (- 2)]}^{2} + 8^{2}} = 10 \\ {(x + 2)}^{2} + 64 = 100 \\ {(x + 2)}^{2} = 36 \\ x + 2 = \sqrt{36} \\ x + 2 = \pm 6 \\ x = 4 or x = - 8 \\ ∴ Q = (4, 8) \end{array}$

$\begin{array}{l} Q T = 10 units \\ T = (4 + 10, 8) \\ T = (14, 8) \\ Gradient of S T = \frac{8 - 0}{14 - (2)} \\ = \frac{8}{16} \\ = \frac{1}{2} \end{array}$

$\begin{array}{l} m_{P R} = m_{S T} = \frac{1}{2} \\ (- 4, 5) : x = - 4, y = 5 \\ y_{1} = m x_{1} + c \\ 5 = \frac{1}{2} (- 4) + c \\ 5 = - 2 + c \\ c = 7 \\ The equation of the straight line P R \\ y = \frac{1}{2} x + 7 \end{array}$

Question 8:
Diagram 4 shows a distance-time graph for the journey of Rina by train from Rawang to Serdang.

Diagram 4

(a)(i) It is given that the speed of the train from Kuang to Serdang is 1.25 km min^-1.
Find the value of t. [2 marks]
(ii) How long it takes, in minutes, to reach Serdang from Kuang? [1 mark]
(b) Calculate the average speed, in km h^-1, of the train for the whole journey. [2 marks]

Solution:
(a)(i)

$\begin{array}{l} Speed = \frac{Distace}{Time} = 1.25 {km min}^{- 1} \\ \frac{(40 - 10) km}{t - 11} = 1.25 \\ \frac{30}{1.25} = t - 11 \\ t - 11 = 24 \\ t = 35 \end{array}$

(a)(ii)
Time taken = t – 11
= 35 – 11
= 24 minutes

(b)

$\begin{array}{l} Average speed = \frac{Total distance}{Totak time} \\ = \frac{40 km}{\frac{35}{60} h} \\ = 68.57 {kmh}^{- 1} \end{array}$

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